First step: Let $I=\langle x-y,xy \rangle \subset \mathbb{C}[x,y]$ be the ideal generated by $u=x-y$ and $v=xy$.
Notice that $x^2 \in I$, since $I \ni xu+v=x(x-y)+xy=x^2-xy+xy=x^2$.
$I$ is not a radical ideal, sicne $x^2 \in I$ but $x \notin I$, hence $I$ is not a maximal ideal.
Notice that the generators of $I$, $u$ and $v$, have only one common zero, namely $(0,0)$; indeed, solving $x-y=0, xy=0$ yields $x=y, x^2=0$, so $x=0=y$.
What is the additional condition (in addition to $u$ and $v$ having only one common zero) that guarantees that $\langle u,v \rangle$ is a maximal ideal?
This seems a basic result in algebraic geometry, see the comments and answers here involving algebraic geometry, in particular:
Known result: "$\langle f,g \rangle$ is maximal if and only if $f=0$ and $g=0$ have exactly one common zero and they're not tangent when they intersect."
Question 1: Where can I find a proof for this quoted result?
Second step: Let $J=\langle f,g,h \rangle$, where $f,g,h \in \mathbb{C}[x,y]$ (still polynomials in two variables).
I wish to find a similar result to the above, something like:
Claim: Assume that no pair of elements of $\{f,g,h\}$ generates a maximal ideal. Then,"$\langle f,g,h \rangle$ is maximal if and only if the curves $f=0,g=0,h=0$ have exactly one common zero and at that common zero, every pair of curves intersect transversally.
Question 2: Is my claim true?
Any help is welcome; thank you very much! Now also asked in MO.